Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include "bits/stdc++.h"

#define int ll
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a)  (a).begin(),(a).end()
#define pb emplace_back
#define INF (1LL<<59)

//verified AOJ GRL_1
#define MAX_V 100000
struct edge{int to,cost;};

void dijkstra(int s, vector<int> &d, vector<edge> G[MAX_V]){
    priority_queue< pii,vector<pii>,greater<pii> > que;
    rep( i,d.size() )d[i]=INF;
    d[s] = 0;
    que.push( pii(0,s) );
    
    while( que.size() ){
        pii p=que.top();
        que.pop();
        
        int v=p.second;
        if(d[v]<p.first)continue;
        
        rep(i,G[v].size()){
            edge e=G[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                que.push(pii(d[e.to],e.to));
            }
        }
    }
}

signed main(){
    int v,e,s,t;
    cin>>v>>e>>s>>t;
    s--,t--;
    
    vector<edge> G[MAX_V];
    
    rep(i,e){
        int a,b;
        cin>>a>>b;
        a--,b--;
        G[a].pb(edge{b,1});
        G[b].pb(edge{a,1});
    }
    
    vector<int> d1(v),d2(v);
    dijkstra(s,d1,G);
    dijkstra(t,d2,G);

    int length = d1[t];
    assert(length!=INF);
    
    vector<int> a(MAX_V,0),b(MAX_V,0);
    for(auto e:d1)if(e!=INF) a[e]++;
    for(auto e:d2)if(e!=INF) b[e]++;
    
    ll ans = 0;
    rep(i,MAX_V){
        int less = length - i - 2;
        if(less<0)continue;
        ans += a[i] * b[less];
    }
    cout<<ans<<endl;
}